Question

Six boys and six girls are to sit in a row at random.The probability that all the six girls and all the six boys sit together is

• A. $\frac{6!6!}{12!}$
• B. $\frac{6!7!}{12!}$
• C. $\frac{2\times 6!6!}{12!}$
• D. $\frac{2\times 6!7!}{12!}$

Answer 1

The correct option is C $\frac{2\times 6!6!}{12!}$

Total ways in which all the 12 people sit : $12!$
Now let's say that all the 6 girls are sitting together and the 6 boys are also sitting together. Hence, each can be treated as an individual group. So, now we have 2 individual groups (1 group of boys and 1 group of girls). Arranging them is $2!$ ways. Also, the internal permutation among the 6 girls is $6!$ and the 6 boys is $6!$
Hence, the answer.